a)đkxđ: \(x+1\ne0\Leftrightarrow x\ne-1\)
\(B=\frac{x^2-x+1}{x^2+2x+1}=\frac{x^2+2x+1-3x}{x^2+2x+1}=1-\frac{3x}{\left(x+1\right)^2}=1-\frac{3\left(x+1\right)-3}{\left(x+1\right)^2}\)
\(B=1-\frac{3}{x+1}+\frac{3}{\left(x+1\right)^2}\)
Đặt \(\frac{1}{x+1}=a\)\(\Rightarrow B=3a^2-3a+1=3\left(a^2-a+\frac{1}{3}\right)=3\left(a^2-2a.\frac{1}{2}+\frac{1}{4}+\frac{1}{12}\right)=3\left(a-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(a-\frac{1}{2}\right)^2\ge0\Leftrightarrow B\ge\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=\frac{1}{2}\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}\Leftrightarrow x+1=2\Leftrightarrow x=1\left(nhận\right)\)
Vậy GTNN của B là \(\frac{1}{4}\)khi \(x=1\)
b) đkxđ \(x-1\ne0\Leftrightarrow x\ne1\)\(E=\frac{3x^2-8x+6}{x^2-2x+1}=\frac{3\left(x^2-2x+1\right)-2x+3}{x^2-2x+1}=3-\frac{2x-3}{\left(x-1\right)^2}=3-\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}\)
\(=3-\frac{2}{x-1}+\frac{1}{\left(x-1\right)^2}\)
Đặt \(\frac{1}{x-1}=b\)\(\Rightarrow E=b^2-2b+3=b^2-2b+1+2=\left(b-1\right)^2+2\)
Vì \(\left(b-1\right)^2\ge0\Leftrightarrow B\ge2\)
Dấu "=" xảy ra khi \(b-1=0\Leftrightarrow b=1\Leftrightarrow\frac{1}{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\left(nhận\right)\)
Vậy GTNN của B là 2 khi x = 2